Multiply the following complex numbers: $({2-i}) \cdot ({4-3i})$
Answer: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({2-i}) \cdot ({4-3i}) = $ $ ({2} \cdot {4}) + ({2} \cdot {-3}i) + ({-1}i \cdot {4}) + ({-1}i \cdot {-3}i) $ Then simplify the terms: $ (8) + (-6i) + (-4i) + (3 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ 8 + (-6 - 4)i + 3i^2 $ After we plug in $i^2 = -1$ , the result becomes $ 8 + (-6 - 4)i - 3 $ The result is simplified: $ (8 - 3) + (-10i) = 5-10i $